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Fifth grade math test questions
Angle APB= angle EPC is known from incident angle = reflection angle.

Because angle APB= angle PAD

So corner pad = corner EPC

Thus, △APD is similar to △PCE in two cases.

The sine of angle c is 4/5.

Do DH perpendicular to BC and H after D.

Then CH=3, DH=4, BH=5, BC=8.

Let BP be x, then

1, angle ADP= angle C.

From the cosine value of the angle c, we can see that the angle ADP is an acute angle at this time, so P is on BH.

So ph = BH-x = 5-X.

According to the sine theorem, we know that the area of a triangle is

1/2×AD×DP×sin angle ADP

= 1/2 × 5 × PD×4/5

=2×PD

In addition, the bottom of the triangular row ADP is AD=5.

The height and DH are equal to 4.

So the area of the triangle ADP is also equal to

5×4/2 = 10

So you get the equation.

2×PD= 10 = >PD=5

Because PD= under the root sign (the square of PH+the square of DH)

DH=4,

So PH=3

So x=5-PH = 2.

Test: When BP=2

It is found that point e will be on the extension line of CD, not on CD.

So this situation does not meet the meaning of the question.

This is the first case.

2, angle APD= angle C.

Pay attention at this time,

Test a situation:

What happens if p and h coincide?

If p and h coincide, we can know the tg angle APD=AD/PD=AD/HD=5/4.

Tg angle C=4/3.

It shows that when they overlap, the angle APD is smaller than the angle C.

Because p moves from left to right,

The angle APD has been decreasing,

This means that point p is still on BH at this time,

Not on HC, although this discussion

It doesn't matter if you don't write, but it is still more important.

Then, continue to calculate:

The ADP area of the triangle is still expressed by base × height /2, that is, 10.

At the same time, it is expressed by sine theorem as follows

1/2 × AP × PD × sin angle APD

= 1/2 × sin angle C × AP × PD

=2/5 × AP × PD = 10

So AP×PD = 25

According to the cosine theorem, there are

Adsquare = apsquare+pdsquare-2cos angle APD×AP×PD.

=AP squared +PD squared -2×AP×PD×3/5

Substituting AP×PD=25, there are

AD square =AP square +PD square -30

therefore

25 = square of AP+square of PD -30

AP square and PD square can be expressed by Pythagorean theorem in BP and PH respectively.

PH=5-BP。

So we can get an equation about BP:

25 = square of BP+16+square of ph+16-30.

therefore

Blood pressure squared +PH squared =23

BP squared +(5-BP) squared =27

Solve the equation with BP=(5+ root number 2 1)/2.

Or BP=(5- radical 2 1)/2.

Check the values of two BP.

All match 0.

But when BP=(5- radical number 2 1)/2, because BP is too small (about equal to 0.2 1), it will make

E is not on the CD, but on the extension line of the CD, which does not meet the meaning of the question and is discarded.

So BP=(5+ radical number 2 1)/2.

Answer over.

The level is limited, please correct me.